As we go down Group 15, the sizes of atoms increase and these are the central atoms in their corresponding hydrides. Shouldn't the large size of central atom (in case of $\ce{Bi}$) actually help the lone pair inside the $\ce{sp^3}$ hybrid orbital to easily be donated and thus increase the basic character down the group?

$\ce{N}$ is small and shouldn't it hold lone pair with much more strength than others, but $\ce{NH3}$ seems to be strong base while $\ce{BiH3}$ is the weakest one.

2 Answers
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Going down this group the proton affinity decreases, so basicity decreases:

$$\mathrm{PA}(\ce{X})=\mathrm{D}(\ce{X+-H})-\mathrm{IE}(\ce{X})$$

where $\mathrm{D}$ is the bond disassociation energy, and $\mathrm{IE}$ is the ionization energy. Down a column this expression is dominated by the $\mathrm{D}(\ce{X^+-H})$ term, and since $\ce{N}$ has the highest bond strength with hydrogen, it is the most basic.

The structural (or hybridization) way to look at it is that the lone pair in $\ce{NH3}$ is very, very different than the lone pair of a similar compound of $\ce{P}$, $\ce{As}$, etc. $\ce{NH3}$ is almost perfectly $\mathrm{sp^3}$ hybridized, but the others have little to no hybridization, so the lone pair is actually held in a orbital that has mainly $\mathrm{s}$ character. This makes the lone pair far less directional (the charge is distributed over a larger area), making it less reactive and harder to form bonds, so less basic.

Finally, $\ce{N}$ is the only one on the series where the HOMO (which contains the lone pair) is concentrated on the $\ce{N}$. For the rest of the series, the HOMO is very delocalized and significant electron density is held on the $\ce{H}$ atoms, so it will not react well as a base as well.

The very idea of $\ce{sp^3}$ hybridization you are referring to is pretty much not there, starting from $\ce{PH3}$. Look at its bond angles, they are almost exactly $90^\circ$. This means that phosphorus contributes to the $\ce{P-H}$ bonds almost exclusively with its $\ce p$ orbitals, and the lone pair just sits there hidden on $\ce s$ orbital, rather than stick out as $\ce{sp^3}$.

The same is even more true for the following hydrides.

Also, with nitrogen being the most electronegative of all, its partial negative charge in $\ce{NH3}$ attracts $\ce{H+}$, which is not the case for $\ce{PH3}$ and the rest.

$\begingroup$As far as I understand the structures , Gr 15 elements do make 4 Sp3 hybrid orbitals. 3 of them make sigma bonds with s orbital of H and one lone pair reside in the 4th sp3 hybrid orbital. You can easily calculate the hybridisation. In case of NH3 , bp-bp repulsion is maximum but as we go down the group electron density shifts away from the central atom due to its large size and thus bp-bp repulsion is reduced but lp-bp repulsion is still the same which makes the H-E-H bond angle closer to 90 degrees. (E is Gr15 elements ). But what does any of these have anything to do with my real question$\endgroup$
– AnindyaOct 28 '15 at 9:06

$\begingroup$Why does the basic character of hydrides go down the group?$\endgroup$
– AnindyaOct 28 '15 at 9:07